Question

Cho 2 đa thức:
P(x)= x² + 2mx + m2
Q(x)= x² + (2m+1)x +m²
Tìm m biết P(1)=Q(-1)

Answer 1

Ta có:

\(P\left(x\right)=x^2+2mx+m^2\)

\(\Leftrightarrow P\left(1\right)=1+2m+m^2\)

\(Q\left(x\right)=x^2+\left(2m+1\right).x+m^2\)

\(\Leftrightarrow Q\left(-1\right)=1-\left(2m+1\right)+m^2=m^2-2m\)

Mà \(P\left(1\right)=Q\left(-1\right)\)

\(\Leftrightarrow1+2m+m^2=m^2-2m\)

\(\Leftrightarrow2m+2m=-1\)

\(\Leftrightarrow4m=-1\)

\(\Leftrightarrow m=\frac{-1}{4}\)

Vậy \(m=\frac{-1}{4}\)